Question

what is the solution to the equation $sqrt3{5x - 4} = sqrt3{7x + 8}$?
$x = -6$
$x = -1$
$x = 1$
$x = 2$

Explanation:

Step1: Cube both sides

To eliminate the cube roots, we cube both sides of the equation \(\sqrt[3]{5x - 4}=\sqrt[3]{7x + 8}\). Cubing a cube root cancels them out, so we get \(5x - 4=7x + 8\).

Step2: Solve for x

Subtract \(5x\) from both sides: \(- 4 = 2x+8\). Then subtract 8 from both sides: \(-12 = 2x\). Divide both sides by 2: \(x=-6\)? Wait, no, wait. Wait, let's check again. Wait, \(5x - 4=7x + 8\). Subtract \(5x\): \(-4 = 2x + 8\). Subtract 8: \(-12=2x\). Then \(x = - 6\)? But let's check the options. Wait, maybe I made a mistake. Wait, no, let's plug in the options. Let's check option A: \(x=-6\). Left side: \(\sqrt[3]{5(-6)-4}=\sqrt[3]{-30 - 4}=\sqrt[3]{-34}\). Right side: \(\sqrt[3]{7(-6)+8}=\sqrt[3]{-42 + 8}=\sqrt[3]{-34}\). Oh, wait, but let's check the other options. Wait, maybe my solving was wrong. Wait, no, when we cube both sides, \((\sqrt[3]{a})^3=a\), so \(5x - 4 = 7x+8\). Then \(5x-7x=8 + 4\), \(-2x=12\), \(x=-6\). But let's check the options. The first option is \(x = - 6\). Wait, but let's check the other options. Wait, maybe the original equation was \(\sqrt[3]{5x - 4}=\sqrt[3]{7x + 8}\)? Wait, maybe I misread. Wait, let's check \(x=-6\): \(5(-6)-4=-34\), \(7(-6)+8=-34\), so cube roots of -34 are equal. So the answer should be \(x=-6\). Wait, but let's check the options again. The first option is \(x=-6\), second \(x = - 1\), third \(x = 1\), fourth \(x = 2\). So according to the solving, \(x=-6\) is correct. But let's check again. Wait, maybe the equation was \(\sqrt[3]{5x - 4}=\sqrt[3]{7x + 8}\), then yes, \(x=-6\) is the solution.

Answer:

A. \(x=-6\)